If $f'(x)=f(x)$ and $f(3)=e^2$, then $f(4)=me^n$ for some integers $m$ and $n$. What are $m$ and $n$ ? $m=~$
Answer: If we let $y=f(x)$ and change the derivative notation, we can rewrite the differential equation so that it's separable. $\dfrac{dy}{dx}=y$ What does it look like after we separate the variables? $\dfrac1y\,dy=dx$ Let's integrate both sides of the equation. $\int\dfrac1y\,dy=\int dx$ What do we get? $\ln|y|=x+C$ What value of $C$ satisfies the initial condition $y(3)=e^2$ ? Let's substitute $x=3$ and $y=e^2$ into the equation and solve for $C$. $\begin{aligned} \ln\,\left|e^2\right| &= 3+C\\ \\ \ln e^2 &= 3+C\\ \\ 2 &= 3+C\\ \\ C&=-1 \end{aligned}$ Now use this value of $C$ to express $y$ in terms of $x$. $\begin{aligned} \ln |y|&=x-1\\ \\ |y|&=e^{x-1}\\ \\ y&=\pm e^{x-1} \end{aligned}$ Which sign satisfies the initial condition $y(3)=e^2$ ? We must choose the positive sign to satisfy the initial condition. $y=e^{x-1}$ If $y(4)=me^n$, what are the integers $m$ and $n$ ? $\begin{aligned} y(4)&=e^{4-1}\\ &=e^3\\ &=me^n \end{aligned}$ Thus, $m=1$ and $n=3$.